+0

# algebra

0
118
1

What is the constant term in the expansion of $$\left(\sqrt{x}+\dfrac5x\right)^{9}$$?

Jul 11, 2020

#1
+8341
0

Considering the general term of the expansion:

$$\left(\sqrt x + \dfrac5x\right)^9 = \displaystyle\sum_{k = 0}^9 \binom{9}k\cdot \left(\sqrt x\right)^k \left(\dfrac5x\right)^{9-k}$$

Simplifying,

$$\left(\sqrt x + \dfrac5x\right)^9 = \displaystyle\sum_{k = 0}^9 \binom{9}k \cdot 5^{9-k}\cdot x^{3k/2-9}$$

When the term is constant, power of x is 0.

$$\dfrac{3k}2 - 9 = 0\\ k = 6$$

Now, substitute k = 6 into the general term, you get the constant term.

(Note that if k was not an integer, the constant term would be 0.)

Jul 11, 2020